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A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of 3.5 revolutions per second. A coin placed at a distance of 1.25 cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is : (g=10 m/s^{2})

  • Option 1)

    0.5

  • Option 2)

    0.3

  • Option 3)

    0.7

  • Option 4)

    0.6

 

Answers (2)

best_answer

As we have learnt

Limiting Friction -

Magnitude of limiting friction between any two bodies in contact is directly proportional to the normal reaction between them.

F_{l}\;\alpha\ R  or   f_{l}=\mu_{s}R

f_{l}= limiting friction 

\mu_{s}= coefficient of friction

R = reaction force

- wherein

*    Maximum value of static friction is limiting friction.

*     Direction is always opposite to relative motion.

 

 

In equilibrium,

m\omega ^{2}r=\mu mg

\mu=\frac{\omega ^{2}r}{g}

\omega=2\Pi v = 7\Pi rad/s

r=1.25cm;g=10

\mu = \frac{49x^{2}.1.25.10^{-2}}{10}=0.6

 

 


Option 1)

0.5

Option 2)

0.3

Option 3)

0.7

Option 4)

0.6

Posted by

prateek

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